Prove or Disprove that Sqrt(2) and Sqrt(4) are IrrationalDate: 23 Jun 1995 08:21:58 -0400 From: Mike Rubitschka Subject: Problems with a mathematics exercise I have learned the following proof that sqrt(2) is irrational: Suppose for contradiction that sqrt(2) is rational. Then sqrt(2) = m/n Assume that m and n are not both even. sqrt(2) = m/n ; m = Sqrt(2)*n and hence m*m = 2*n*n This shows that m*m is an even number. Thus, m itself is even. This means m = 2*p for some integer p. But then 4*p*p = m*m = 2*n*n or n*n = 2*p*p This shows that n*n is even, and so it follows that n is even. But that contradicts the fact that not both m and n are even ... So sqrt(2) must be irrational. Problem: If you use this proof with sqrt(4) you can proof that sqrt(4) is irrational!!! : Suppose for contradiction that sqrt(4) is rational. Then sqrt(4) = m/n Assume that m and n are not both divisible by 4. sqrt(4) = m/n ; m = Sqrt(4)*n and hence m*m = 4*n*n This shows that m*m is divisible by 4. Thus, m itself is div. by 4. This means m = 4*p for some integer p. But then 16*p*p = m*m = 4*n*n or n*n = 4*p*p This shows that n*n is div by 4, and so it follows that n is div. by 4. But that contradicts the fact that not both m and n are div. by 4 ... So sqrt(4) must be irrational??? Where is the fault which I must have done? Since I could not find it, would you please help me? Greetings and thanks in advance. Michael Rubitschka Vienna, Austria Date: 23 Jun 1995 09:34:59 -0400 From: Dr. Ethan Subject: Re: Problems with a mathematics exercise Hey Mike, Well, That is a great question. and if your method was right we would have quite a problem. But here is the problem. You say: " This shows that m*m is divisible by 4. Thus, m itself is div. by 4." Well that isn't true. What if m is 2 then 2*2 is divisible by 4 but 2 isn't. Does that make sense? Write back if we can help you in any way. Ethan - Doctor On Call |
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